Question

面试题 53 - I. 在排序数组中查找数字 I

题目描述

统计一个数字在排序数组中出现的次数。

 

示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: 2

示例 2:

输入: nums = [5,7,7,8,8,10], target = 6
输出: 0

 

提示：

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums 是一个非递减数组
• -109 <= target <= 109

 

注意：本题与主站 34 题相同（仅返回值不同）：https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/

解法

方法一：二分查找

由于数组 nums 已排好序，我们可以使用二分查找的方法找到数组中第一个大于等于 target 的元素的下标 $l$，以及第一个大于 target 的元素的下标 $r$，那么 target 的个数就是 $r - l$。

时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。其中 $n$ 为数组的长度。

class Solution:
  def search(self, nums: List[int], target: int) -> int:
    l = bisect_left(nums, target)
    r = bisect_right(nums, target)
    return r - l

class Solution {
  public int search(int[] nums, int target) {
    int l = lowerBound(nums, target);
    int r = lowerBound(nums, target + 1);
    return r - l;
  }

  private int lowerBound(int[] nums, int x) {
    int l = 0, r = nums.length;
    while (l < r) {
      int mid = (l + r) >>> 1;
      if (nums[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  }
}

class Solution {
public:
  int search(vector<int>& nums, int target) {
    auto l = lower_bound(nums.begin(), nums.end(), target);
    auto r = upper_bound(nums.begin(), nums.end(), target);
    return r - l;
  }
};

func search(nums []int, target int) int {
  l := sort.Search(len(nums), func(i int) bool { return nums[i] >= target })
  r := sort.Search(len(nums), func(i int) bool { return nums[i] > target })
  return r - l
}

impl Solution {
  pub fn search(nums: Vec<i32>, target: i32) -> i32 {
    let search = |x| {
      let mut l = 0;
      let mut r = nums.len();
      while l < r {
        let mid = l + (r - l) / 2;
        if nums[mid] >= x {
          r = mid;
        } else {
          l = mid + 1;
        }
      }
      l as i32
    };
    search(target + 1) - search(target)
  }
}

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function (nums, target) {
  const search = x => {
    let l = 0;
    let r = nums.length;
    while (l < r) {
      const mid = (l + r) >> 1;
      if (nums[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  };
  const l = search(target);
  const r = search(target + 1);
  return r - l;
};

public class Solution {
  public int Search(int[] nums, int target) {
    int l = search(nums, target);
    int r = search(nums, target + 1);
    return r - l;
  }

  private int search(int[] nums, int x) {
    int l = 0, r = nums.Length;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (nums[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  }
}